Solved Problem on Dynamics
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A locomotive, mass 130 t, drags a railroad car, mass 120 t. The maximum force that the locomotive coupler can withstand is 2,900 kN. Find the maximum driving force that the locomotive can exert in order not to break the coupler. Disregard resistance forces.


Problem Data:
  • Mass of the locomotive:    mL = 130 t = 130,000 kg;
  • Mass of the railroad car:    mR = 120 t = 120,000 kg;
  • Maximum force supported by the coupler:    T = 2,900 kN = 2,900,000 N.
Problem diagram:

The system can be represented by two blocks, representing the locomotive and the railroad car connected by a rope representing the coupling between the two blocks (Figure 1).

Figure 1

We choose a reference frame oriented to the right, in the same direction as the force \( {\vec F}_{\small M} \) that produces an acceleration \( \vec a \) to the train.
Figure 2

Drawing a Free-Body Diagram for each body.

  • Railroad car (Figure 3):
    • Horizontal direction:
      • \( \vec T \): tension force at the coupler.
    • Vertical direction:
      • \( {\vec N}_{\small R} \): normal reaction force on the railroad car;
      • \( {\vec W}_{\small R} \): weight of the railroad car.
Figure 3

  • Locomotive (Figure 4):
    • Horizontal direction:
      • \( {\vec F}_{\small M} \): driving force exerted by the locomotive;
      • \( -\vec T \): tension force at the coupler.
    • Vertical direction:
      • \( {\vec N}_{\small L} \): normal reaction force on the locomotive;
      • \( {\vec W}_{\small L} \): weight of the locomotive.
Figure 4

Solution

Applying Newton's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec F=m\vec a} \tag{I} \end{gather} \]
  • For the railroad car:
In the vertical direction, there is no motion, normal force and weight cancel each other out.
In the horizontal direction
\[ \begin{gather} T=m_{\small R}a \tag{II} \end{gather} \]
  • For the locomotive:
In the vertical direction, there is no motion, normal force and weight cancel each other out.
In the horizontal direction
\[ \begin{gather} F_{\small M}-T=m_{\small L}a \tag{III} \end{gather} \]
Equations (II) and (III) can be written as a system of two equations with two unknowns (FM and a)
\[ \begin{gather} \left\{ \begin{array}{r} T=m_{\small R}a\\ F_{\small M}-T=m_{\small L}a \end{array} \right. \end{gather} \]
we solve the first equation of the system for acceleration a
\[ \begin{gather} a=\frac{T}{m_{\small V}} \end{gather} \]
substituting the acceleration into the second equation of the system
\[ \begin{gather} F_{\small M}-T=m_{\small L}\frac{T}{m_{\small V}} \end{gather} \]
\[ \begin{gather} F_{\small M}=T+m_{\small L}\frac{T}{m_{\small V}}\\[5pt] F_{\small M}=(2900000\;\mathrm N)+(130000\;\mathrm{\cancel{kg}})\times \frac{(2900000\;\mathrm N))}{120000\;\mathrm{\cancel{kg}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{\small M}=6,041,666\;\mathrm N} \end{gather} \]

Note: since the acceleration found is unthinkable for a train
\[ \begin{gather} a=\frac{T}{m_{\small V}}\\[5pt] a=\frac{2,900,000\;\mathrm N}{120,000\;\mathrm{kg}}\\[5pt] a=24.2\;\mathrm{m/s^2} \end{gather} \]
it would accelerate from 0 to 100 km/h in approximately 1 second. The coupling of a train is very strong, it must be subjected to a force too large to break.
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